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# PHYS 200: Fundamentals of Physics I

## Lecture 15

## - Four-Vector in Relativity

### Overview

The discussion of four-vector in relativity continues but this time the focus is on the energy-momentum of a particle. The invariance of the energy-momentum four-vector is due to the fact that rest mass of a particle is invariant under coordinate transformations.

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html## Fundamentals of Physics I## PHYS 200 - Lecture 15 - Four-Vector in Relativity## Chapter 1: Recap: The Four-Vectors of Position, Velocity and Momentum in Space-Time [00:00:00]
Let me remind you of what we were trying to do. We had learned that space-time is described by four coordinates, so that if you’re following some particle or some event, you label it with four numbers x, _{1}x, _{2}x_{3}. Now, I’m of two minds about x and _{2}x; sometimes I carry them around, sometimes I don’t. _{3}x is just our familiar _{1}x; x is _{2}y and x is _{3}z. It’s possible to arrange all the motion to take place in the x direction, in which case y and z don’t change when you go from one frame to the other, so I don’t usually bother with that. But sometimes I put it back because if you have only two components, x and _{0}x, you might say, “Hey, the world is two dimensional.” So, I want to keep reminding you now we think of it terms of four dimensions. So, at least for that reason sometimes they carry all the three components of position, sometimes only one. But I’m sure you guys can figure out from the context which is which._{1}So, going back to the simpler situation of only x - _{1}βx, divided by the square root. And _{0}x′ will be _{0}x times _{0} - βx, divided by that square root, where _{1}β is v/c. Sometimes it’s useful to introduce γ, which is [reciprocal of square root of ] 1 - u^{2}/c^{2}. Also to prevent all confusion, u will be the velocity of your frame relative to mine. v will stand for the velocity of some particle that we are both studying. So, let’s reserve the symbol v for motion of actual particles, which is why we are there. We are all observers but we’re observing something. That something has a velocity that I’ll call v and you will call w, and u is the speed relative to your frame and my frame.All right, so this is the notion of space-time because space and time are getting mixed up when you go to a new frame of reference. But even in this world where nothing is sacred, the following combination is very important, which we denote as r. r dot r will be x. So, you can use either notation. Whenever possible I just keep two, but you’ve got to remember, this is why it’s a four-dimensional dot product. And this is the same for all people. So, I can calculate it, you can calculate it, our ^{2} + y^{2} + z^{2}x and _{0}x will differ, but it’s the property of the Lorentz transformation, that if I square this guy and subtract it from the square of this guy, on the right-hand side some magical things will happen and I’ll get the same combination without the primes. That’s like the length of the vector. And that’s called the space-time interval. Let’s call that _{1}s, the space-time interval.^{2}If it took two events separated in space by xy, the other is the rotated xy. If you go to the rotated frame, the components x and y change, but how far you are from the origin, or what’s the length of the vector is the same for all people. More generally, the dot product of two vectors is invariant under rotations, and this is the new dot product. I use the symbol dot product, but I don’t draw arrows, because if I don’t draw any arrows and use capital letters, it means it’s a four-vector. That’s the X vector.Then I said, “Let’s try to reason by analogy and try to invent a momentum vector.” How did we invent the momentum vector in the old days? We said, let the particle move a distance I wanted the same thing now in this present problem, but I now want to define obviously a four-dimensional vector. So, to do the four-dimensional vector, my plan is to define something called momentum, which is now going to be the change in this fourth position divided by something. The question is, “What do I divide it by?” I cannot divide by x. It moves the distance _{1} Δx in a time _{1} Δt, and we know that for this piece of its motion the following quantity, Δs, which is [root of] c Δt, is the same for all people. So, I should really be dividing by this.^{2} - Δx^{2}Let me write it slightly differently as c. So, we are asked to divide by this crazy thing. Well, let me write this crazy thing as ^{2}c times something I want to call dτ, is dt times the square root of 1 minus Δx over Δt times 1/^{2}c. Then I said, “We can give a meaning to this thing.” In fact, let’s take the derivative with respect to ^{2}τ not s. They differ only by this factor c, which is a constant so it doesn’t matter. τ has units of time, and you can ask, “What does that time stand for?” It’s not the time according to me, but I argued that it’s the time according to the particle. Why? Because you can evaluate this quantity in any frame you like. So, go to the frame moving with the particle. If you’re moving with the particle, and if you’re here now and there later, later is certainly later, but here remains here for the particle. It doesn’t move in its own frame. So, Δx will vanish, therefore Δs simply reduces to c times the time elapsed according to the particle. So, dτ is the time elapsed according to the particle itself. According to the particle. That’s what we’re going to divide by.So, it’s a very clever way to circumvent the question of what should I take the derivative respect to, my time and your time all are variable, we don’t agree on it. But let’s ask how much time elapsed according to the one thing we are studying, and let’s divide by that time. That answer will be the same because you and I may not agree on how long it took to go from here to there, but we can all ask how long did the particle say it took to go from here to there. We’re talking about the same thing, and we’ll get the same answer. To get the particle time you’ve got to take the time according to you and me and multiply it by this factor. This factor, by the way, is nothing but usual time dilatation. Because the particle says it took time So, let’s rewrite this in a way that’s going to be more helpful to us. Take the ratio and write p. The four-momentum p is going to be m times dx times _{0}/dτmdx. I’m not going to bother with the other components. You can always put them back. But _{1}/dτx you remember was _{0}ct. So, this really becomes mdt divided by dτ, mdx divided dτ. This is guaranteed to be a four-vector. That means, if you know the components of this in one frame of reference, and you call one of them as p, and you call this one as _{0}p. Then, in a frame that’s moving relative to your _{1}x speed u, our p′ will be _{0}p minus _{0}β times p_{1} over the square root, and p′ will be _{1}p minus _{1}β times p over the square root. It will transform just the components of _{0}x. Yes?
c go?
ct, you’re quite right. It is c times dt. You need that; then, this is correct. So, we’ve got a four-vector, but we don’t know what it means. Because we don’t know what dτ means. Of course it’s the time according to the particle, but I don’t care about the particle. I’m studying this particle, I’ve got my own clock, I’ve got my own meter sticks, I want to think in terms of time according to me. Period. Well, we can do that. We can do that by going to anything where you take a derivative such as dx and you can write is as _{1}/dτdx times _{1/}dtdt/dτ. dx is something we understand, it’s just the velocity according to you and me. If you use that rule here, you will find the momentum has got components _{1}dtm times c, dt/dτ is dt/dt, which is 1, times dt divided by dτ, which is this factor here. And the second thing will be mv divided by the same thing. In other words, you can trade τ derivatives for t derivatives everywhere, provided you attach this factor. That’s the ratio of the τ derivative to the t derivative.So, now I have a four-vector. I still don’t know what it means, but at least it is written in terms of things with a direct operational significance for me. I see a particle moving, it’s got a velocity, put that in and I get this thing. Apparently, the relativity theory says this should play the role of momentum, but a four dimensional momentum in your new world. If you want to bring the other components, we’ll put an arrow, you can keep it or leave it; it doesn’t matter. Now, notice once again that–Okay, I’ll continue some more. I’ve already done this last time, but let me remind you. We looked at this guy and said, “Look, at smaller velocities, when mc times 1 plus v over 2^{2}c plus dot dot dot. That becomes ^{2}mc + 1/2mv times 1 over ^{2}c plus dot dot dot. And now I finally have something I understand, because ½ mv is familiar kinetic energy. But I don’t like this annoying ^{2}c here. If I multiply both sides by c, I learn that c times p is _{0}mc plus ½ ^{2}mv, plus more stuff that’s more and more powers of ^{2}v over c. But this I argue. If that’s the kinetic energy, everything is an energy formula, so the whole thing stands for energy. So, I learned that the first component, of P is energy divided by c. So, I come back here and I write this as E/c.## Chapter 2: The Energy-Momentum Four-Vector [00:15:53]So, for all of you, or for many of you who found some of this mathematics difficult to follow, it depends. For me it was love at first sight. This is really when I said, “I want to go into business.” For some reason this worked for me. But it doesn’t matter. What you have to know is what I’m going to tell you now. Operationally, I’m not going to hold you responsible for this part of the material at all. For me, it’s part of explaining how logically starting with the very first axiom–first two axioms you can deduce everything. But now, I want to tell you operationally what you guys are supposed to know. You are supposed to learn the following two things. In the old days, we had time and we had space; they have been united into a single quantity; ( x etcetera. In other words, the three components of a vector joined with one quantity that looked like a scalar together to form a four-dimensional object. Now, we are learning that there is another four-dimensional object, one of which is just the momentum, and the other is energy divided by _{1}c. And we want to call that as p _{0},p etcetera. Again, the familiar momentum, which is a vector, joined with another quantity, which the best way to think about it is kinetic energy, plus rest energy, plus other corrections. But it’s a scalar; it doesn’t depend on orientation of axis in the old days. But now under Lorentz transformation, namely, when you go from one frame to another, _{1}E and p will all mix with each other just like x and t will mix with each other, but this is the new four-vector. It’s called the energy-momentum four-vector. Just like that’s called the space-time four-vector. This is the energy-momentum four-vector. So, all over relativity you will find three objects, which used to be part of a vector, will combine with the fourth one, which used to be a scalar; together, they’ll form a new four-dimensional vector.So, we should know the following. When we are studying particle dynamics, we have to know in more detail–in fact, let me dedicate a whole board to that, it’s very important. The new definition is that it’s
mv plus other stuff, right? So, I recognize ½ ^{2}mv as an energy; then, the other powers ^{2}v, they’re all corrections obviously to the energy at higher velocity. Just like momentum, even in relativistic theory was not ^{4}/c^{4}mv. It started out like mv, but when you expand the denominators and powers to v/c you get more and more corrections. I identify this as an energy because if I go to very low velocities, it reduces the rest mass, which is never going to change, plus a ½ mv, which played the role of kinetic energy.^{2}That’s why it has units of energy, and one term in it looks like energy. So, the whole thing, the theory is telling you, it’s all just the energy of the particle. In the old days when velocities were very small, we saw only that two terms are written: ½ mc. ^{2}mc we didn’t bother with because it doesn’t change in any collision. You add the ^{2}mc before, each particle has got its own ^{2}mc before and after; they simply cancel. So, the ½ ^{2}mv squares before and after that would equal, and we use that to solve collisions. Now we will be told you should use the conservation of this energy and the conservation of this momentum in a collision.Okay, so the first thing is this is v, and ^{2}/c^{2}p in one dimension is mv divided by 1 minus v/^{2}c. This you must know. So, momentum and relativity looks a bit like the old momentum but it’s got this ubiquitous factor. And energy looks really weird; doesn’t look like anything on Earth, but if you expand this in powers of ^{2}v, you will see the terms we wrote down. Rest energy plus everything else you can call energy of motion, and the first term of the energy of motion is the familiar ½ mv. In other words, this is how the world always was. It was revealed to us in the old days when we only looked at low energy particles as ^{2}mc + ½ ^{2}mv. And we couldn’t see the dot dot dot terms because ^{2}v/c was so small, the fourth power and sixth power and eighth power of v/c were too small for us to detect.Likewise, the momentum of a particle was never For that purpose, you should take the energy and divide by x, that will behave like _{0}x, under Lorentz transformation. But there are other consequences of the fact that _{1}P is a four-vector. The first consequence it this: you remember the dot product of a vector with itself is the same for all observers. So, what’s the dot product of P with itself?Well, before I do that, let me remind you in general, in relativity you will have four-vectors: (A) you’ll have one part that looks like time, and one part that looks like an ordinary vector. And (B), is another four-vector, it’s got a part that looks time and another part that looks like an ordinary vector. It’s a union up of former vector with a former scalar. And they will transform under Lorentz transformation just like t. But the following quantity A dot B, which is A minus the usual dot product of the spatial components, is invariant. That means all people will agree on this quantity. And a special case is _{0} B_{0}A dot A, which equals A minus the length of the vector part, which for our purpose will be just _{0}^{2}A - _{0}^{2}A. This is all reasoning by analogy. Yes?_{1}^{2}
Okay. So, now let’s ask what is the invariant length square of the momentum four-vector with itself. That means, you’ve got to do m–I’m sorry, you’re going to take _{0}mc over this factor square, minus mv over this factor square. Well, let’s try to do this in our head, guys. m is going to come out of the whole thing, you agree? On the top, I have got 1 over 1 minus ^{2}v over ^{2}c, yet I have minus ^{2}v over 1 minus ^{2}v over ^{2}c, so they will all cancel giving you ^{2}mc. Uh, ^{2}m.^{2}c^{2}[Note: Here, the Professor has started dropping the subscript on When you take a velocity vector for a particle, it doesn’t have a fixed length. The particle can be moving slow, it can be moving fast. So I want to find mc, there is no velocity for you, and you have no momentum. As the particle has no momentum as seen by the particle so the components of the vector P become mc and 0. What’s the length square of this vector? Well, it’s the square of this guy and case closed; there’s nothing here to subtract from here. That’s a quick way to verify that this length of vector with itself is in fact m. In other words, the trick is ride with the particle and calculate the vector. Then, the vector has only one component, namely ^{2}c^{2}mc. It doesn’t have a spatial part because it’s not moving according to itself. That’s where P dot P is m. So, people write it also as follows: remember ^{2}c^{2}p was _{0}E over c, this is just P is ^{2}m. This is one way to write it explicitly, but here is how people write it: multiply both sides by ^{2}c^{2}c and write ^{2}E is equal to ^{2}c ^{2}p^{2}- m; that deserves a box. That’s how we remember it. I have just rearranged this equation. ^{2}c^{4}P dot P is m. By definition ^{2}c^{2}P dot P is the time-like component square minus the spatial component square. We agree the time-like part is E/c; the spatial part is just the momentum. So, I’ve just rearranged it to get this. And it says the energy of the particle depends on its momentum, if it comes complete to rest, you kill this term, that’s E = mc; that’s the rest energy of a particle.^{2}Now, I am almost finished with the kinematics except for one thing. There are strange particles in the world called photons. A photon has no rest mass; therefore, if you naively went by any of these formulas, if you killed the mass of the photon, you might think energy and momentum vanish because c times p)^{2} is what a photon obeys. So, photons can have energy, and photons can have momentum, but their energy will be c times their momentum so that if you find the invariant square of the photon momentum you will get 0. So, photons are particles whose momentum square is 0. But the square is the four-dimensional square with the time square minus the spatial square.So, let me summarize once more for people who may not want to follow every detail. The practical part of what you must understand from what I’ve been teaching you is that when you study a single material particle, which has a mass, so that it can be brought to rest and made to sit on the table, we associate with that at four-vector. The first part of that is the energy over mc divided by this and mv divided by this. And one can explicitly verify that P dot P is m. This is a summary of all that I’ve said. If you don’t know anything at all, know this. Yes?^{2}c^{2}
K dot K, is 0 for photons. A photon is a particle whose square of the four-momentum is 0. Material particles that can come to rest have a square of the fourth momentum, which is m. This is something you have to know.^{2}c^{2}Another thing you have to know is that when I write them in this notation as four-vectors, you should know the following: if you know the components, p in one frame, and you move relative to _{1}S with speed u, p′ and _{0}p′ will be related to _{1}p and _{0}p by the Lorentz transformation. Just like _{1}x′ and _{0}x′ are related to _{1}x and _{1}x by that formula on the top left. I’m going to do a lot of problems for you so you get used to doing relativistic kinematics.Last thing I want to tell you is why do we take this definition of momentum. Okay, you agree with the theory naturally gave out something that it said deserves to be called momentum. But what’s so nice about this? Why is it important that the momentum-energy form a vector? I’m going to give you a reason for it, and try to follow the reason. The whole advantage of energy and momentum–What is the big deal about energy and momentum, let me ask you that. Yes?
## Chapter 3: Relativistic Collisions [00:32:20]
Let’s understand why. Suppose you’ve got a collision; a lot of particles come in; one particle brings momentum-energy P. And at the end of the collision maybe they become particle 3 and particle 4 and particle 5. In relativistic theories, if two particles come in, you can have 36 particles coming out. In fact, that’s why they built accelerators. If two went in and two came out, it’s a waste of time. You can smash any two things and with that energy you can produce matter. So, it’s not a mistake I’ve made. I do mean that sometimes two can go into three, or four or five. Let’s take an example where this is true._{2}Each P′, will also equal _{2}P′ plus _{3}P′ plus _{4}P′. That’s guaranteed, and I want to tell you why. Suppose these were ordinary vectors, and let me make life simple by saying that _{5}P became _{1} + P_{2}P. What that means is one vector, _{3}P, another vector, _{1}P, added up to one vector, _{2}P. Except these are not ordinary arrows, these are arrows in four dimensions. The fact that two arrows add up to a third one, are equal to a third one, means they form a triangle. Now, if they can form a triangle for me, and you can stand on your head and do what you like, it will still be true. Even though it looks like this, it is still true that this guy plus this guy is equal to that guy. If two vectors add up to a third one; namely, if a sum of two vectors is equal to another vector; namely, the left-hand side vector is equal to the right-hand side vector, going to a rotated frame will continue to say that the rotated vector is equal to the rotated vector. That’s because equal vectors upon rotation will remain equal vectors. That’s the whole point. The total incoming momentum four-vector, call it _{3}P incoming, and this is the outgoing momentum, call it the P outgoing. If P incoming is equal to P outgoing for me, so two vectors that equal in space-time, for the rotated observer they will be equal.Do you understand why? Because my total incoming vector will undergo Lorentz transformation to become your incoming vector. My outgoing vector will undergo Lorentz transformation to find your outgoing. But if the input of the Lorentz transformation is the same, the output will also be the same because you will be putting the very same numbers to get out the output numbers. That’s why it’s very important that this momentum that we have be used, because it has a chance that if it’s conserved in one frame, it’s conserved in all frames. If you went and said, “You know what? I don’t like your formulas. I’m really wedded to this formula for momentum,” what you will find is that even if it’s conserved in your frame of reference, in a different frame of reference, that combination, sum over So, now I’m going to do sample problems to give you a feeling for that. So, as usual, they will start out with the simplest one and then try to make it more complicated. So, the first problem is this. I’m standing here, I’ve got some mass So anyway, the question here is, “What is P as the final one. Let’s write down all the equations we can; the only equation in town is _{2}K + P. Everybody is a four-vector, so don’t be fooled. It’s four equations. But in this problem–;luckily, since there is no motion perpendicular to the _{1} = P_{2}x axis, we can have this x momentum; there’s no need for y and z. Because the photon comes this way, and I can start moving this way, but all the action is in the x direction. So, here it will be actually two-dimensional vectors and not four. There’s nothing vertically before or after, it will be 0 = 0. So, what are those equations? Before you do that, let’s write down the momentum of the photon first. Momentum of the photon has got some energy [should have said p] _{0}/c, and it’s got some momentum k, where = kc, you’ve got to remember that. P has got an energy which is _{1}mc, and no spatial momentum because I’m not moving. At the end of the day P, I am moving, but I also have a new mass, and I have a new velocity, which we don’t know, and I have _{2}m′ v divided by the square root of 1 minus v/^{2}c. Now,all of you should be able to write this part down. If you’re not, you should take a minute, you should think about it, you can stop me, you can question me. But if you cannot do this, you cannot do any of the problems. You’ve got to think, “Would I have done this? Does it make sense to me?” The bookkeeping is all in the top. When you’ve got a material particle, it’s first term, 0 component, will be ^{2}m times c divided by the square root of all those things. But if the particle is at rest, you only have mc. There is no momentum. The final object, me–After I got hit by the photon, I am recoiling; so, I will have a velocity. I don’t know what it is, but my energy in terms of that, the first term, p will be this; this will be _{0}P._{1}## Chapter 4: Law of Conservation of Energy and Momentum Using the Energy-Momentum Four-Vector [00:41:07]Now, the Law of Conservation of Energy and Momentum, namely /c plus mc, will be equal to m′ c over this. Then, k plus 0 will be equal to m′ v divided by this. This is not any different from the old days. You had Law of Conservation of Momentum, you had Law of Conservation of Energy, except energy is not ½ mv; energy is this crazy thing. Plus, even if a particle is not moving, don’t forget to give it ^{2}mc as the energy; that’s the new part. So, what is the question I asked you? I said, “What is m′.” So, your job is to take these two simultaneous equations, this one and this one, and solve for m′. So, what should I do? Can you think of a strategy that will give me m′ ?
m′ times c, and that you get by taking the square of this minus the square of this. You can juggle that, but I’m not going to do it that way, I’m going to show you a much more efficient and rapid way to do this.^{2}So, let’s do it the more efficient way, so let me come down here so you can see. Here is a shortcut. The shortcut is to work entirely with four-vectors and four-dimensional dot products, and don’t get down to the ugly business of writing each component. We know that P plus _{1}K, right? And who are we after? We’re after m′; then we remember that P, namely _{2}^{2}P dot _{2}P is equal to _{2}m′ square c. Yes? So, the question you are asked is–No one asked you how fast I am moving after I got hit by the photon; you’re only asking me what is my new rest mass. See, I’ve swallowed a photon, so I weigh more; it’s really that. How much more do I weigh is the question; namely; what is my rest mass. Namely, I swallowed it, and I recoiled, and if you travel with me and find me at rest, what mass ^{2}m′ will you measure? So, no one’s asking me my velocity of recoil v. So, you shouldn’t bother to calculate that. You could if you wanted to, but no one is asking you that. So, go directly for P. So, the quantity we’re looking for is _{2}^{2}m′ square c. That’s got to be ^{2}P, the whole thing square. Well, do the old algebra. That’s equal to _{1} + KP plus twice _{1}^{2} + K^{2}P dot _{1}K. Okay, now let’s see if I can do what we do next. Would you like to guess? Yes? No? Do we know any of these quantities? Do we know anything about the three terms I’ve written here? Yes?
K dot K is 0. You don’t have to worry about that. How about the square of the momentum of me? Well, that was my rest mass before, times c plus 0. Now, ^{2}P dot _{1}K you’ve got to be careful. That dot product means–let me write down the vector of P and _{1}K for you. P was _{1}mc and 0, and K was /c, and the spatial momentum, where was equal to kc. Remember that? For a photon, once you pick the energy, you have no choice with the momentum. The magnitude of the momentum has to be the magnitude of energy divided by c. So, let’s take P dot _{1}K; let’s try to do P dot _{1}K. Remember the dot product of two vectors is time times time. mc times over c minus space dot space, that was 0. So, I cancel that, and I find it’s really m. And that’s equal to m′ c. Yes?^{2}
P and _{2}P sub-0 and _{1}K sub-0, too many indices. So, what we’re trying to do is whenever we write a _{1}K whose components are and little k, we know we’re talking about a photon. And whenever I write a P, I know I’m talking about the electron. So, is just the energy of the photon. So now, if I cancel all the c squares, then I get m′ square plus m plus 2^{2}m/c. Now, you can take the square root, if you like, and that’s the answer.^{2}So, my mass would have gone up. You might think, “How much would the mass have gone up?” First of all, this guy swallowed a photon of energy , so maybe that energy over
Okay, so this is one problem. I was going to do one or two more, but I want to make sure you guys are following; I just don’t know. See, if I ask you what you would do next, if you have some idea, then we can proceed, but if you don’t know, maybe we should slow down. And you’ve got to talk about which part of it you find more difficult. So, let me say a few things. And the first time you hear something that you find difficult, you’ve got to put your hand up, okay? Otherwise, you’re not going to make progress. The first thing I say is this: particles of mass have an energy and momentum that depend on their velocity and rest mass as given by the first equation. And the energy and momentum are assembled into a four-vector, the first component of which is m for any particle of mass independent of its state of motion. And the simplest way to remember that is to evaluate the dot product in the frame moving with the particle. In that frame, it only has a ^{2}c^{2}p, which is _{0}mc, and doesn’t have a spatial momentum; so that’s got to be the answer.Second thing, when you deal with photons, they’re a little tricky. Photons have energy, photons have momentum, but energy is not mass times some velocity square over 2 etcetera. They have no mass, but they have energy and they have momentum, and the condition of the photon is that the energy must be momentum times c times P)^{2} + m. For a photon, there is no ^{2} c^{4}m. That means E = pc. So, photon energy and photon momentum have relative magnitude. Yes?
k as _{0}/c, then it means = kc. That’s the second statement. That’s not obvious from anything I’ve taught you. So, that’s some new stuff. Because suddenly, I’ve got an object with no mass and the whole derivation for momentum started with m times some velocity and so on, so that fails for photons. But they exist and they carry energy, they carry momentum. You should think of them as particles that obey E, with ^{2} = (cp)^{2} + m^{2}c^{4}m = 0, so they obey E = pc. That’s a second statement.Third statement is, in a collision you add up all the That’s it. That’s the trick to collisions, balancing energy and momentum in one mega equation, which is four-momentum before equals four-momentum after. And the beauty is any observer can look at this collision from any frame of reference; different people will disagree on what’s the energy and what’s the momentum of the particle. For example, in this example I gave you, I said I am at rest, and I got hit by a photon, then I’m moving. Well, that’s the point of view of some observers for whom I was at rest initially. You can see the whole thing from a rocket. In that case, my initial momentum will have a spatial part because on the rocket, I’ll be moving backwards. This photon will also have a new and a new
Now I want to do a second very interesting problem, which is this. The year is 1953 or ‘54. You are trying to build a collider in Berkeley. It’s going to have a proton target. By the way, I’m going to show you one more trick that’s very useful. Put m. I already know that if that’s the mass square and that’s the mass squared. Well, that’s the mass, and that’s an energy, so I know that should be a c here. You can always fudge the ^{2}c at the end of the day, there’s no reason to carry it from the beginning. Just from dimensional analysis, when you’re looking for energy and you see a mass, put a ^{2};c there. When you’re looking for a mass and you see an energy and it should be a mass, divide by ^{2}c. So, that’s what I’m going to do.^{2}So, here is a proton at rest. Here is a proton that I’ve created from my Bevatron, which is going to come and smash into this. And the job of this reaction is the following: where this notation is very unfortunate [
v over ^{2}c should be 3 times ^{2}mc. That’s the energy. Because that 3^{2}m plus this m will make 4m. Yes?
So, the trick is again going to be the same. So, this is the initial state, this is the final state, but this final state has got some motion. It cannot be at rest. So, what’s the initial momentum? Initial momentum for this guy is some energy
Now, let me finish this and maybe we’ll have a discussion. So, E + m) - ^{2}P = 16^{2}m. Now, so, let’s see where to write this; maybe I’ll write it here. That means ^{2}E 2^{2} + m^{2} +Em - p = 16^{2}m. But you must know that ^{2}E is an ^{2} - p^{2}m. I’m saying ^{2}E, if you banish all factors of ^{2} = p^{2} + m^{2}c. So, this ^{2}E is an ^{2} - p^{2}m; there is already an ^{2}m here, so I get ^{2}2m 2^{2} +Em = 16m. You cancel the 2 and make it 14^{2}m, and you divide by 2^{2}m, you find E = 7m. By the way, this is posted on the web, on the internets. So, you guys can read it whenever you like. So, don’t worry about this last detail; I’ve given you every possible detail here. If you didn’t get every factor, it doesn’t matter. What I want you to understand is, I’ve also given you the whole argument on how you go to this frame and that frame. The point is you can answer E = 7m, now we realize it’s time to put back the velocity of light, and everybody knows, even man and woman on the street, that this should be really 7mc. If we carried factors of ^{2}c throughout the calculation, it would have been very messy, but here is where c would have to turn up. So, if you’re looking for an energy and you get a mass, you put back the ^{2}c.^{2}So, what we are told is, the incoming proton, as it moves, its energy goes from v over ^{2}c is vanishing. Wait until the energy is 7 times the rest energy, and when it’s 7 times the rest energy, that’s when it’s ready to hit the static proton and produce proton/anti-proton pairs from nowhere. So, some of the 7^{2}m is going to give a momentum to the fragment of the collision, and some is to create them. And the way we found out the minimum energy is we did energy-momentum calculation. But the minimum energy configuration is easily imagined in the center of mass frame, that’s the whole point. Not in the lab frame, the center of mass frame. That’s why accelerators, nowadays, are taking one particle that can move faster and faster and hitting a target they produce intersecting storage rings where the particles are going in opposite directions. And every time they meet in the racetrack, they’re allowed to collide. What’s the beauty of that? The beauty of that is, since they come with opposite momenta, the final fragments don’t have to move at all. So, you can make more and more mass. So, don’t put your energy all into one beam and a static target; put them into two oppositely moving beams. Then, all the energy is available for producing mass.^{2}All right, so I’ll stop now. I think this is a difficult chapter, so you have to go and read, and you have to talk to each other and try to do as many problems as you can. [end of transcript] Back to Top |
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